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Alie Ibarra C16 #59
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Alie Ibarra C16 #59
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# Defines a node in the singly linked list | ||
from requests import head | ||
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class Node: | ||
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def __init__(self, value, next_node = None): | ||
def __init__(self, value, next_node = None, prev_node = None): | ||
self.value = value | ||
self.next = next_node | ||
self.previous = prev_node | ||
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# Defines the singly linked list | ||
class LinkedList: | ||
def __init__(self): | ||
self.head = None # keep the head private. Not accessible outside this class | ||
self.head = None # keep the head private. Not accessible outside this class | ||
self.tail = None | ||
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#---------------------- | ||
# returns the value in the first node | ||
# returns None if the list is empty | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(1) | ||
# Space Complexity: O(1) | ||
def get_first(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
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if self.head: | ||
return self.head.value | ||
return None | ||
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#---------------------- | ||
# method to add a new node with the specific data value in the linked list | ||
# insert the new node at the beginning of the linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(1) | ||
# Space Complexity: O(1) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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def add_first(self, value): | ||
pass | ||
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if self.head is None: | ||
self.head = self.tail = Node(value) | ||
else: | ||
new_node = Node(value, self.head) | ||
self.head.previous = new_node | ||
self.head = new_node | ||
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#---------------------- | ||
# method to find if the linked list contains a node with specified value | ||
# returns true if found, false otherwise | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def search(self, value): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
current = self.head | ||
if current == None: | ||
return False | ||
while current: | ||
if current.value == value: | ||
return True | ||
current = current.next | ||
return False | ||
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#---------------------- | ||
# method that returns the length of the singly linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def length(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
counter = 0 | ||
current = self.head | ||
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while current != None: | ||
current = current.next | ||
counter +=1 | ||
return counter | ||
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#---------------------- | ||
# method that returns the value at a given index in the linked list | ||
# index count starts at 0 | ||
# returns None if there are fewer nodes in the linked list than the index value | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def get_at_index(self, index): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
counter = 0 | ||
current = self.head | ||
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while current != None: | ||
if counter == index: | ||
return current.value | ||
counter += 1 | ||
current = current.next | ||
return None | ||
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#---------------------- | ||
# method that returns the value of the last node in the linked list | ||
# returns None if the linked list is empty | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def get_last(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ However, you could also just access the tail's value since this is a doubly linked list and achieve O(1) time complexity |
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pass | ||
current = self.head | ||
if current == None: | ||
return None | ||
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while current.next != None: | ||
current = current.next | ||
return current.value | ||
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#---------------------- | ||
# method that inserts a given value as a new last node in the linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def add_last(self, value): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ However, again, since you have a |
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pass | ||
if self.head == None: | ||
self.add_first(value) | ||
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else: | ||
new_node = Node(value) | ||
current = self.head | ||
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while current.next: | ||
current = current.next | ||
current.next = new_node | ||
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#---------------------- | ||
# method to return the max value in the linked list | ||
# returns the data value and not the node | ||
def find_max(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
max_value = 0 | ||
if self.head == None: | ||
return None | ||
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current = self.head | ||
while current: | ||
if current.value > max_value: | ||
max_value = current.value | ||
current = current.next | ||
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return max_value | ||
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#---------------------- | ||
# method to delete the first node found with specified value | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def delete(self, value): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
current = self.head | ||
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if current == None: | ||
return | ||
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while current: | ||
if current.value == value: | ||
if current == self.head: | ||
self.head = current.next | ||
self.head.previous = None | ||
return | ||
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if current == self.tail: | ||
self.tail = current.previous | ||
self.tail.next = None | ||
return | ||
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current.previous.next = current.next | ||
current.next.previous = current.previous | ||
return | ||
current = current.next | ||
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#---------------------- | ||
# method to print all the values in the linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def visit(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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helper_list = [] | ||
current = self.head | ||
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@@ -84,34 +171,52 @@ def visit(self): | |
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print(", ".join(helper_list)) | ||
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#---------------------- | ||
# method to reverse the singly linked list | ||
# note: the nodes should be moved and not just the values in the nodes | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def reverse(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
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## Advanced/ Exercises | ||
if not self.head: | ||
return | ||
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current_node = self.head | ||
previous_node = None | ||
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while current_node: | ||
next = current_node.next | ||
current_node.next = previous_node | ||
previous_node = current_node | ||
current_node = next | ||
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self.head = previous_node | ||
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#---------------------- Advanced/ Exercises ----------------------- | ||
# returns the value at the middle element in the singly linked list | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def find_middle_value(self): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
mid_value = int(self.length() / 2) | ||
return self.get_at_index(mid_value) | ||
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#---------------------- | ||
# find the nth node from the end and return its value | ||
# assume indexing starts at 0 while counting to n | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def find_nth_from_end(self, n): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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pass | ||
nth_from_end = self.length() - (n+1) | ||
return self.get_at_index(nth_from_end) | ||
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#---------------------- | ||
# checks if the linked list has a cycle. A cycle exists if any node in the | ||
# linked list links to a node already visited. | ||
# returns true if a cycle is found, false otherwise. | ||
# Time Complexity: ? | ||
# Space Complexity: ? | ||
def has_cycle(self): | ||
pass | ||
return False | ||
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# Helper method for tests | ||
# Creates a cycle in the linked list for testing purposes | ||
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😃 Nice! Doubly Linked List!